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**Particles Behaving as Waves**

Chapter 39 Particles Behaving as Waves

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**Goals for Chapter 39 To study the wave nature of electrons**

To examine the evidence for the nuclear model of the atom To understand the ideas of atomic energy levels and the Bohr model of the hydrogen atom To learn the fundamental physics of how lasers operate To see how the ideas of photons and atomic energy levels explain the continuous spectrum of light emitted by a blackbody To see how the Heisenberg uncertainty principle applies to the behavior of particles

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Introduction At the end of the 19th century light was regarded as a wave and matter as a collection of particles. Just as light was found to have particle characteristics (photons), matter proved to have wave characteristics. The wave nature of matter allows us to use electrons to make images (such as the one shown here of viruses on a bacterium).

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De Broglie waves The physicist de Broglie hypothesized, from an argument for symmetry in Physics, that the relationships E = hf = hc/ and p = h/ for photons also apply to electrons. Thus electrons should have wave characteristics. The de Broglie hypothesis was confirmed by the discovery by Bell Labs physicists Davisson and Germer in 1927 that electrons undergo diffraction when interacting with crystalline matter, just like x rays do (see Figure 39.4). Note: Because electrons have mass and do not travel at the speed of light, they DO NOT obey the relation lf = c. Kinetic energy for non-relativistic particles can be written If we accelerate an electron through a potential difference V, it gains a kinetic energy eV, so we can write: Thus, the wavelength of an electron accelerated through potential V is Diffraction, as usual, obeys

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Electron Diffraction Example 39.1: In an electron-diffraction experiment, an electron is accelerated by a potential difference of 54 V and a maximum in the diffraction pattern occurs at 50°. X-ray diffraction shows the atomic spacing in the target to be d = 2.1810-10 m. Find the electron wavelength. Alternatively, we could use the diffraction information: Accelerating to a higher energy would yield a smaller wavelength for the electron, which suggests using electrons a probes for imaging at high resolution.

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Electron microscopy Example 39.3—An electron microscope. One can use magnets as focusing elements, and create an “optical” system for electrons. Note, the blue cones in the diagram are not ray paths like we used for geometric optics. What voltage is needed to accelerate electrons to a wavelength of 10 pm (roughly 50,000 times smaller than visible light)?

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Atomic line spectra As we saw last week, if He gas is sealed in a glass tube and heated to form a hot gas, the light emitted by the He atoms includes only certain discrete wavelengths. The spectrum of this light (“line spectrum”) is different for different elements. Nineteenth-century physics had no explanation for this.

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The nuclear atom Rutherford probed the structure of the atom by sending alpha particles at a thin gold foil. Some alpha particles were scattered by large angles, leading him to conclude that the atom’s positive charge is concentrated in a nucleus at its center. Example 39.4.

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**Size of a Gold Nucleus Example 39.4: Rutherford Scattering**

An alpha particle (charge 2e) is aimed directly at a gold nucleus (charge 79e). What is the minimum kinetic energy of the alpha particle in order to approach within 5.0x10-14 m of the center of the gold nucleus? Assume the gold nucleus remains at rest. The alpha particle must overcome the electrostatic repulsion of the gold nucleus, which can be characterized by the potential energy at the point of closest approach By energy conservation, the incoming kinetic energy must be at least this high. By changing the incoming kinetic energy, the atom can be probed at different depths, which allows the size of the nucleus to be determined.

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**The failure of classical physics**

Rutherford’s experiment suggested that electrons orbit around the nucleus like a miniature solar system. However, classical physics predicts that an orbiting electron would emit electromagnetic radiation and fall into the nucleus (see Figure 39.14). So classical physics could not explain why atoms are stable. Niels Bohr explained atomic line spectra and the stability of atoms by postulating that atoms can only be in certain discrete energy levels, and cannot further decay from a minimum energy called the ground level. The atom is said to be in the ground state.

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Atomic energy levels When an atom makes a transition from one energy level to a lower level, it emits a photon whose energy equals that lost by the atom (see Figure at lower left). An atom can also absorb a photon, provided the photon energy equals the difference between two energy levels (see Figure at lower right). When an atom is in a higher energy level, it is said to be in an excited state. Energy levels for soduim

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The same gas that, when heated, gives emission lines at certain wavelengths will also show the same spectrum in absorption (i.e. it is able to remove photons from the incoming light by absorbing them and raising the energy level of the atom to an excited state).

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**The Bohr model of hydrogen**

Bohr explained the line spectrum of hydrogen (see Figure below) with a model in which the single hydrogen electron can only be in certain definite orbits. In the nth allowed orbit, the electron has quantized orbital angular momentum nh/2. Recall that angular momentum is For a circular orbit, the magnitude of momentum for the nth orbit is (note that this means h has units of momentum, check it!). Bohr’s argument is rather complicated, but with it he was able to explain quantitatively the spectral lines of hydrogen.

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**Classical Explanation for Quantized Momentum**

One way to get a (rather misleading) physical intuition about the quantized nature of atomic orbits, consider the wavelength of an electron in various orbits. It makes sense that electrons, due to their wave nature, ought to require orbits that are an integral number of wavelengths, as if they were standing waves. If we require then rearrangement gives Bohr’s result Also, since the electrostatic force must supply the centripetal force to keep the electron in its circular orbit, we have Combining these two equations allows us to solve simultaneously for rn and vn:

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**Scale Drawing of Orbits and Standing Waves**

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**Hydrogen spectrum in more detail**

The line spectrum at the bottom of the slide is not the entire spectrum of hydrogen; it’s just the visible-light portion, involving transitions to the n = 2 energy state. Here, n is the principle quantum number. Hydrogen also has series of spectral lines in the infrared and the ultraviolet.

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**Hydrogen-like atoms Numerical value of the Rydberg constant is:**

If a higher-charge nucleus has only a single electron, the same pattern of energy levels exists, but one must take account of the nuclear charge Z: Note that all bound energies are negative, and approach zero as n gets larger. Here are the energy levels for hydrogen (H) and singly-ionized helium (He+). The simple picture does NOT work for atoms with more than one electron! Note, we have not taken account of the finite mass of the nucleus. To be correct, we should use the reduced mass in place of the electron mass

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**Distribution of States**

Absorption of radiation vs. collisional excitation. Spontaneous emission vs. stimulated emission. Atoms spontaneously emit photons of frequency f when they transition from an excited energy level to a lower level. Excited atoms can be stimulated to emit coherently if they are illuminated with light of the same frequency f. This happens in a laser (Light Amplification by Stimulated Emission of Radiation). One can relate the temperature of a gas to the distribution of energy states in the gas using the Boltzmann equation. The ratio of atoms in state n to atoms in the ground state is given by

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**Example: Hydrogen Lines in the Solar Spectrum**

Say we heat hydrogen to 6000 K (the temperature of the Sun’s surface). What is the ratio of atoms in the first excited state (n = 2) compared to the ground state (n = 1)? The value of (E2 – E1) is The value of kT is: Thus This is a very small number, but nevertheless the Sun has deep absorption lines in the Balmer series (due to transitions from even higher energy states).

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**Continuous spectra and blackbody radiation**

A blackbody is an idealized case of a hot, dense object. Such objects show a continuous spectrum rather than spectral lines. The reason is that many-many photon interactions scatter and shift the wavelengths, spreading them into a characteristic shape as shown at right, called the blackbody spectrum. The three characteristics of a blackbody spectrum are: The intensity grows with temperature as I = sT 4, where s = 5.6710-8 W m-2 K-4 is the Stefan-Boltzmann constant. The peak shifts according to lmT = 2.90 10-3 m-K (called the Wien displacement law) The shape of the distribution I(l) is invariant (after scaling for the above two factors). Classical physics could not explain the shape of the blackbody spectrum. Rayleigh had an idea that explained the lower-frequency part: Consider normal modes of waves in a box, with energy equipartion kT.

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**Continuous spectra and blackbody radiation**

Rayleigh’s formula works well for long wavelengths, but “blows up” at short wavelengths, which was referred to as the ultraviolet catastrophe. Planck provided an explanation by assuming that atoms in the blackbody have evenly-spaced energy levels, so that at shorter wavelengths (higher energies) there were fewer “states” for the photons to be in. Planck’s formula was entirely empirical: This formula explains all three of the characteristics of the blackbody spectrum in the previous slide. It was this empirical quantization of photon energy that Einstein showed was actually true with his explanation of the photoelectric effect.

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**Continuous spectra and blackbody radiation**

Example 39.7: Light from the Sun To a good approximation, the Sun is a blackbody with temperature 5800 K. (a) At what wavelength does the Sun emit most strongly? To answer this, we use the Wien displacement law (b) What is the total radiated power per unit surface area? For this, we use the Stefan-Boltzmann law Example 39.8: A slice of sunlight Find the power per unit area radiated from the Sun’s surface in the wavelength range to nm. We need an integral over the Planck curve. Because this is such a narrow range, we can approximate by assuming the intensity is nearly constant: Plugging in all of the numbers, we get I = 0.39 MW/m2.

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**The Heisenberg Uncertainty Principle revisited**

The Heisenberg uncertainty principle for momentum and position applies to electrons and other matter, as does the uncertainty principle for energy and time. This gives insight into the structure of atoms. Example 39.9: The uncertainty principle: position and momentum. Consider an electron bound to the ground-state orbit, According to the Heisenberg uncertainty principle, the momentum uncertainty is This is associated with an energy uncertainty: Example 39.10: The uncertainty principle: energy and time. A sodium atom remains in its first excited state for about 1.610-8 s before decaying to the ground state and emitting a photon of l = 589 nm. What is the uncertainty in energy of that state? What is the spread of wavelengths of the photon emitted?

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